Notice that $F$ is the mansion point.

please anyone has the correct version?

First let's show that F is on the triangle ABC circumscribed circle. Let F be the intersection of AE with the triangle ABC circumference, and it is enough to show that (D, F, K) is a collinear point. <OAB = <CAK <-> <OAF = <EAK <-> <ADF = (<AOF / 2) = (<AEK / 2) = <ADK CF=IF=EF, IE=2IF=2CF, TRI CIL~TRI EIB ->CI/CL=IE/BE -->CI/CL=2CF/BE --> BE*CI=2CF*CL [END]

번역에 약간 이상한 감이 없지 않아서 말씀드립니다. 첫 문장에 triangle with distinct sides라고 하셨는데 부등변 삼각형임을 말하고 싶으면 distinct side length라고 하셔야되요. 직접적으로 부등변 삼각형을 표현하자면 scalene triangle이라고도 쓸 수 있습니다.

그러게... 어떻게 보면 정삼각형도 "triangle with distinct sides"라고 할수있네요 KMO에서 공식으로 주는 영어 버전 없나요? 문법이 어색해서 KMO 문제 푸는게 불편합니다.

Letteer wrote:

Dude I think you posted the solution for the day 2 not day 1 LOL. Anyways, here is my solution. WLOG, let $AB<AC.$ Also, let $\angle A =2\alpha, \angle B =2\beta, \angle C =2\gamma.$ Consider the expression $\frac{BE\cdot CI}{2CL}.$ Since, $(I,B,EC)$ are concyclic, $\angle BEI=\gamma$.Thus, $BE=IE\cos{\gamma}$. Also, we have $CI=\frac{CL}{\cos{\gamma}}.$ Plugging those lengths into our expression above, we have $\frac{BE\cdot CI}{2CL}=\frac{IE}{2}=IM=CM$ where $M$ is the intersection of $AI$ and $\Omega.$ Therefore, it is enough to show that $(D,M,K)$ are collinear. $\angle ADM=\angle ADB +\angle BDF =\alpha+2\gamma$. $\angle ADK=\frac{1}{2}\angle AEK=\frac{1}{2}(180^{\circ}-2\angle KAE)=90^{\circ}-\angle KAE=90^{\circ}-(\angle BAF-\angle BAK)$ $=90^{\circ}-(\alpha-(90^{\circ}-2\beta))=\alpha+2\gamma$. Hence, $\angle ADM=\angle ADK$ and points $D,M,K$ are collinear.

Let $P$ be the intersection of segments $AE,\ BC$. Let $F'$ be the midpoint of arc $BC$ opposite to $A$. Since point $E$ is the center of circle $ADK$ we have $$\measuredangle ADF=\measuredangle ADK=90^\circ-\measuredangle KAE=\measuredangle APB=\measuredangle ACF'=\measuredangle ADF'$$Because $F,F'\in AE$ above equality proves $$F=F'$$. Let $Q$ be the midpoint of segment $CI$. By trilium theorem $FI=CF$ thus $\angle FQC=90^\circ$. Let $M$ be the point satisfying $EM\perp BC$ and $M\in BC$. Then $$\angle EBM=90^\circ-\frac{\angle ABC}{2}=\frac{\angle CAB+\angle BCA}{2}=\angle BCF+\angle BCI=\angle ECI=\angle ECQ$$Hence (a,a,a) $$\triangle BME\sim\triangle CQF\implies \frac{BM}{BE}=\frac{CQ}{CF}=\frac{CI}{2CF}$$The rest is simple $$BM=\frac{BC+AC-AB}{2}=CL.$$QED

Let $F'=(ABC)\cap\overline{AI}.$ Notice that \begin{align*}\measuredangle ADF'&=\measuredangle ABF'\\&=\measuredangle ABC+\measuredangle CBF'\\&=\measuredangle ABC+\measuredangle CAX\\&=\measuredangle ABC+\measuredangle XAB\\&=\measuredangle AXB\\&=90+\measuredangle XAK\\&=\tfrac{1}{2}\measuredangle AEK\\&=\measuredangle ADK\end{align*}since $BICE$ is cyclic with center $F'$ by the Incenter-Excenter Lemma. Thus, $F'$ lies on $\overline{DK}$ and $F'=F.$ Notice that $$\measuredangle LCI=\measuredangle ICB=\measuredangle IEB$$and $\angle EBI=\angle CLI=90$ so $\triangle BIE\sim\triangle LIC.$ Hence, $LC/BE=IC/IE$ and $2LC\cdot FC=BE\cdot IC.$ $\square$